How to find first term, common difference, and sum of an arithmetic progression? Given that an arithmetic progression is such that the 8th term is twice the second term, and the 11th term is 18.
Find:
1) The first term and common difference.
2) The sum of the first 26 terms.
3) The smallest of the progression whose values exceed 126?
How on earth am I meant to solve this? I'm guessing you try and find a formula for the nth term, but I have no clue how to get there.
Any suggestions?
 A: First we write out our given information:
$$a_8=2a_2$$
$$a_{11}=18$$
$a_n$ is an arithmetic sequence. 
Where here $a_n$  means the $n$th term of our sequence.
What does an arithmetic sequence mean? It means to get to the next term in your sequence you add a constant ($c$) each time:
$$a_{n+1}=a_n+c$$
Equivalently:
$$\frac{a_{n+1}-a_{n}}{(n+1)-n}=c$$
So $a_n$ is of slope $c$ ($c_2$ is another constant):
$$a_n=cn+c_2$$
Where here $c_2=a_0$ (Substitute in $n=0$ and see why that has to be the case if we let $a_0$ exist)
Now we use the other given information to try to come up with a solution.
Let $n=2$:
$$a_2=2c+c_2 {}{}$$
Let $n=8$, using the above equation we have:
$$a_8=8c+c_2=2a_2=4c+2c_2 {}{}{}{}$$
Let $n=11$
$$a_{11}=18$$
$$a_{11}=11c+c_2$$
But $a_{11}-a_8=(11c+c_2)-(8c+c_2)=3c$
Hence, $a_{11}=3c+a_{8}$
$$a_{11}=3c+4c+2c_2=18$$
$$a_{11}=3c+8c+c_2=18$$
Solve this system of equations to get a closed form for the arithmetic sequence.

$$a_n=1.2n+4.8$$.
 We can check it works $a_2=1.2(2)+4.8=7.2$. Now we compute $a_8$ to see if $a_8=2a_2$ as required: $a_8=1.2(8)+4.8=14.4=2(7.2)=2a_2$. It is arithmetic as we may check $a_{n+1}-a_n$ is a constant $1.2$. Also $a_{11}=1.2(11)+4.8=18$ as required.

The answers follow from this, from summation formulas/methods of evaluating sums, and from algebra.
A: From $$\begin{cases}a+7d=2(a+d),\\a+10d=18\end{cases}$$
$$d=\frac65,a=6.$$

$$S=26\frac{6+\left(6+25\,\dfrac65\right)}2.$$

$$6+(n-1)\frac65>126\implies n\ge102.$$
